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Ç
ENEE 324 Engineering Probability Lecture 2
Counting
For the case of rolling a single fair die, let Ω = 1, 2, 3, 4, 5, 6 and let
A = 2Ω = ∅, Ω, 1, 2, 3, 4, 5, 6, 1, 2, 2, 3...There are 26 = 64 members in A, and only 6 of these are elementary events.Declare that all singletons (elementary events) are equally likely. [recall fair-ness assumption]
Since P (Ω) = 1 and Ω is the union of 6 singletons, all equally likely, it followsfrom the addition axiom that,
Probability of a singleton = 16.
The probabilities of all the other events in A can be determined from thisone fact ! We simply apply the addition axiom.
In a deck of well-shuffled cards, the probability of drawing the heart = 152
.
Example 1: Toss a coin repeatedly until the first head. In each toss,
P H = p : P T = 1 − p = q.
Ω = 1, 2, 3, ... = sample space of # tosses needed until first head.Assume p 6= 0.
pj = p j tosses until first head= qj−1 · p j = 1, 2, ....
Where did this come from?
1
∞∑
j=1
qj−1p = p · (1 + q + q2 + ...)
= p · limn→∞
(1 + q + q2 + ... + qn)
= p · lim
n → ∞1 − qn
1 − q
=p
p(because q < 1)
= 1
Example 2: Lifetime of computer memory chip satisfies: “proportion ofchips whose lifetime exceeds t decreases exponentially at the rate α.”Here α > 0.
Ω = (0,∞)P [(t,∞)] = e−αt t > 0
P [(0,∞)] = e−0α = 1 as it should be.
P [(r, s)] = P [(r,∞)] − P [(s,∞)] = e−αr − e−αs , r < s
Some combinatorics
(a) Given n distinct things, how many ways can we permute them?Think of this as filling n marked cells
1 2 3 . . . n
Fill cell 1 in any of n ways.Fill cell 2 in any of (n − 1) ways (with the remaining (n − 1) things).Fill cell 3 with any of (n − 2) ways.
2
Fill cell n in (1) way.
Total number of ways of filling cells is nPn = n(n − 1)(n − 2) · · · 3 · 2 · 1We call this n!.
(b) Given n distinct things, how many different permutations of r thingscan we make from these n things?
Treat the problem as one of filling r out of n cells.
Proceeding as before we get
nPr = n(n − 1)(n − 2)...(n − r + 1)
=n!
(n − r)!
SAMPLING(c) Given n distinct things, how many combinations of r things out of thesen things can we make?Denote this yet to be determined quantity as nCr.Combinations ignore order. Thus,
nCr · r! = nPr
=n!
(n − r)!·
Hence nCr =n!
(n − r)!r!
The sampling is said to be random if all of these combinations are equally
3
likely. So the probability of a particular combination being picked up in arandom sample is (n−r)!r!
n!
It is common to use the notation(
nr
)
instead of nCr. These integers have along history. Newton’s binomial expansion says
(a + b)n =n∑
k=0
(
n
k
)
akbn−k
Proof: (a+ b)n is an expression, homogeneous of degree n. Hence each term
in (a + b)n will be of the form akbn−k. How many are of this form?
(
n
k
)
2
Identitites
(i)
(
n
r
)
=
(
n
n − r
)
(ii)
(
n
r
)
=
(
n − 1
r − 1
)
+
(
n − 1
r
)
Single out an object – ak, say.Numbers of choices of r objects out of n objects = (number of choices that
exclude ak) + (number of choices that include ak) =(
n−1r
)
+(
n−1r−1
)
Example 3: (sample without replacement)
Total of N items.Choose n at random without replacement.This will yield
(
Nn
)
possible samples.
If the N items are made up of r1 blues and r2 reds, r1 + r2 = N , then theprobability of choosing exactly s1 blues and (n − s1) reds, (here s1 ≤ n and
4
s1 ≤ r1, (n − s1) ≤ r2), is given by
(
r1
s1
)(
r2
n−s1
)
(
Nn
)
We call this the hypergeometric law
Where did this come from?Answer: Think of each sample as equally likely and count how many thereare favorable to the event of interest.
Example 4: (inspection for quality control) A batch of 100 manufactureditems is checked by an inspector, who examines 10 items selected at random.If none of the 10 items is defective, the batch of 100 is accepted. Otherwise,the batch is subject to further inspection. What is the probability that abatch containing 10 defectives is accepted?
Solution: Number of ways of selecting 10 items of a batch of 100 is
N =
(
100
10
)
.
All such samples are equally likely.
A = event that the batch is accepted by the inspector. Then A occurs if all10 items of the selected sample belong to the set of 90 non-defectives.
Number of combinations (samples) favorable to A is:
N(A) =
(
90
10
)
P (A) =N(A)
N
=
(
9010
)
(
10010
) =90!
10!80!
10!90!
100!
≈ (1 − 1
10)10 ≈ 1
e2
5
Example 5: What is the probability that two cards picked randomly froma full deck are aces?
Solution
N = 52 cardsn = 4 aces.
There are(
522
)
equally likely picks.
N(A) =(
42
)
ways are favorable to getting 2 aces.
P (A) =N(A)
N=
(
42
)
(
522
) =6 · 2
52 · 51=
6
26 · 51=
1
2212
Theorem: Given a population of n elements, let n1, n2, ...nk be positive in-tegers such that n1 + n2 + ...nk = n. Then there are precisely
N =n!
n1!n2!...nk!
ways of partitioning the population into k sub-populations of the prescribedsizes and order.
Proof: Order of sub-populations matters.
(n1 = 4, n2 = 2, n3, ..., nk) 6= (n1 = 2, n2 = 4, n3, ..., nk).
Order within sub-populations does not matter.
6
N =
(
n
n1
)(
n − n1
n2
)(
n − n1 − n2
n3
)
· · ·(
n −∑k−2i=1 ni
nk−1
)
=n!
n1!(n − n1)!
(n − n1)!
n2!(n − n1 − n2)!
(n − n1 − n2)!
n3!(n − n1 − n2 − n3)!
· · · (n −∑k−2i=1 ni)!
nk−1!nk!
=n!
n1!n2!...nk!2
Example 6: What is the probability that each of 4 bridge players holds anace?
n = 52n1 = n2 = n3 = n4 = 13
From the theorem, there are n!n1!n2!n3!n4!
equally likely deals.
There are 4! = 24 ways of giving an ace to each player.
Remaining 48 cards can be dealt in 48!12!12!12!12!
ways.
Thus these are 24 · 48!(12!)4
distinct deals favorable to the desired event.
P (event) = 24 ·48!
(12!)4
52!(13!)4
≈ 0.105
Use Stirling’s formula to get this approximation.
7
Stirling’s Formula (following Feller)Let an = n!
(n)nn = 1, 2, ...
an+1
an=
(n + 1)!
(n + 1)n+1
/ n!
(n)n
=(n + 1)n!
(n + 1)n(n + 1)
(n)n
n!
=1
(
1 + 1n
)n
Let bn = n!(
en
)n= anen
loge
bn+1
bn= 1 + loge
an+1
an
= 1 − n loge
(
1 +1
n
)
= 1 − n
(
1
n− 1
2n2+
1
3n3− · · ·
)
=1
2n− 1
3n2+ · · ·
Let βn = n!(
en
)n+ 1
2
loge
βn+1
βn= 1 −
(
n +1
2
)
loge
(
1 +1
n
)
= − 1
12n2+
1
12n3− · · · < 0.
Hence βn+1 < βn. We have shown that βn is a montone decreasing se-quence which is bounded below by 0. Thus β = limβn
n→∞exists.
8
In other words,
βn = n!(
e
n
)n+ 1
2 → β a constant
So we can take n! ∼ β · (n)n+ 1
2 e−(n+1/2). Verify that β =√
2πe.
Thus
n! ∼√
2π(n)n+ 1
2 e−n (I)
There is a slightly better one.
n! ∼√
2π(n)n+ 1
2 e−n+ 1
12n (II)
Formulas (I) and (II) are respectively the first and second approximations ofStirling.
9
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ENEE 324 Engineering Probability Lecture 4
Applications of Bayes’ Theorem
Example: There are 10 urns, 9 of which are of type I and 1 of type II. Urnof type I carries 2 white balls and 2 black balls. Urn of type II carries 5 whiteballs and 1 black ball.
If a ball drawn randomly from a randomly chosen urn turns out to be white,then what is the probability that the chosen urn is of type II? This is a modelof an inference problem.
Solution
A := ball drawn is whiteB1 := urn is of type IB2 := urn is of type IIB1 and B2 are disjoint events and define a partition Ω = B1 ∪ B2.
P (B2|A) = P (A|B2) P (B2)P (A|B1) P (B1)+P (A|B2) P (B2)
P (B1) = 9/10 ; P (B2) = 1/10 Prior probabilitiesP (A|B1) = 2/4 = 1/2P (A|B2) = 5/6
P (B2|A) =5/6 · 1/10
1/2 · 9/10 + 5/6 · 1/10
=5
27 + 5=
5
322
1
Statistical Independence; The idea that two phenomena have nothing todo with each other has a key role in probability theory.Definition We say that in an experiment E , two events A and B are statis-
tically independent if,
P (A ∩ B) = P (A) · P (B)Imagine a long series of trials, each of which involves carrying out two ex-periments E1 and E2, where only E1 leads to A1 and only E2 leads to A2.
If n = total number of trials, n(A1 ∩ A2) = number of trials leading tooccurence of A1 and A2, then
P (A1 ∩ A2) ∼ n(A1 ∩ A2)
n
P (A2) ∼ n(A2)
n
P (A1) ∼ n(A1)
n.
On the other hand
P (A1 ∩ A2) ∼ n(A1 ∩ A2)
n
=n(A1 ∩ A2)
n(A2)· n(A2)
n
∼ P (A1) · P (A2)
The following example illustrates statistical independence and related sub-tleties. Throw two dice resulting in the outcomes (X, Y ).Let A1 : event that X is odd
A2 : event that Y is oddA3 : event that X + Y is odd.
2
Clearly, A1 and A2 are independent.
P (A1) = 12
= P (A2)P (A3) = Prob X odd and Y even
+Prob X even and Y odd= 1
2· 1
2+ 1
2· 1
2
= 12
P (A3|A1) = Prob Y even= 1
2
P (A3|A2) = Prob X even= 1
2
⇒ P (A3|A1) = P (A3) = P (A3|A2)
Thus A3 and A1 are independent and A3 and A2 are independent. 2
Definition: Given events A1, A2, . . . , An, we say these are mutually indepen-
dent if:
P (Ai ∩ Aj) = P (Ai) · P (Aj)
P (Ai ∩ Aj ∩ Ak) = P (Ai)P (Aj)P (Ak)
...
P (A1 ∩ A2 ∩ · · · ∩ An) = P (A1)P (A2) · · ·P (An).
In the previous example, the events A1, A2, A3 are not mutually independent,even though they are pairwise independent, because
P (A1 ∩ A2 ∩ A3) = 0
but
P (A1)P (A2)P (A3) =(
12
)3.
3
Probability Trees: When an experiment is of a sequential nature, it is oftenconvenient, especially for purposes of calculation, to represent the experimentgraphically by a probability tree. It is a rooted tree and the vertices representoutcomes/events of the experiment. The edges are labelled by the conditional
probabilities required to descend from a given vertex to an adjacent one. Theprobability associated with the event corresponding to a vertex is obtainedby taking under consideration the product of the probabilities labelling theedges forming the unique path between the vertex, and the root of the tree.
Example: Flipping a coin three times:
∅
P(H ) P(T )
H T
P(H |H ) P(T |H ) P(H |T ) P(T |T )
H H H T T H T T
P(T |H H ) P(T |H T ) P(T |T H ) P(T |T T )
H H H H H T H T H H T T T H H T H T T T H T T T
1 1
1 1
2 1 2 1 2 1 2 1
1 2 1 2 1 2 1 2
3 21 3 21 3 21 3 21
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
fig. 1
4
Corresponding Pascal’s Triangle
1 1
1 12
1 13 3 fig. 2
Probability trees may also be infinite. We give an example below.
Example: Player A flips a fair coin. If the outcome is a head, he wins; if theoutcome is a tail, player B flips. If B’s flip is a head, he wins; if not, playerA flips the coin again. This process is repeated (ad infinitum, if necessary)until somebody wins. What is the probability that A wins?
5
A
B
A
A
A
A
A
B
B
B
B
B
T
T
TT
TT
TTT
TTT
H
H
TH
TH
TTH
TTH
etc
fig. 3
For the probability tree above, the darkened vertices correspond to the ele-
mentary events for which A wins. Since the probability represented by eachbranch of the tree is 1/2, we have:
PA wins calculated via sampling with replacement
6
= PAH + PATH + PATTH + · · ·
=1
2+(
1
2
)3
+(
1
2
)5
+ · · ·
=1
2
[
1 +(
1
2
)2
+(
1
2
)4
+(
1
2
)6
+ · · ·]
=1
2
1
1 −(
12
)2
=1
2
1
1 − 1/4
=2
3
There is a big advantage for A to flip first.
Gambler’s Ruin – (Application of Total Probability Law)
Example: (1) Toss coin. Call correctly, win 1 dollar. Call wrongly, loose 1dollar.
Payoff Matrix
@@
@@
@@
Tail −1 1
Head 1 −1
TailHeadCall
Toss
Fig. 4
7
Initial Capital = x dollars and x is a positive integer.
STRATEGY PLAY UNTIL EITHER :ւց
Win m Dollars Lose Shirt(i.e. has a total ( RUIN)of m dollars)
Question: What is the probability p(x) of ruin?
A = RUINB1 = Win first call = p
B2 = Lose first call = (1 − p)
P (A) = P (A|B1) P (B1) + P (A|B2) · P (B2)p(x) = p(x + 1) · 1
2+ p(x − 1) 1
21 ≤ x ≤ m − 1
= p(x + 1) p + p(x − 1) · (1 − p)
B.C.
p(0)p(m)
==
10p(x) = C1 + C2x is the solutionC1 = 1 C1 + C2m = 0
Hence:
p(x) = 1 − x/m 0 ≤ x ≤ m
If p 6= 1/2 the solution is not linear
Example [Matching]:
n distinct items to be matched against n distinct cells. What is the proba-bility of at least 1 match?
Solution:
Ak := event that kth item is matched (we don’t care about the rest)P (n) = Probability of at least 1 match
8
= P (∪nk=1Ak)
=n∑
i=1
P (Ai) −n∑
i<j=2
P (Ai ∩ Aj)
+n∑
i<j<k=3
P (Ai ∩ Aj ∩ Ak · · ·)
+(−1)n+1P (A1 ∩ A2 ∩ · · ·An)
= P1 − P2 + P3 · · · ± Pn
P (Ai1 ∩ Ai2 · · · ∩ Aim) =(n − m)!
n!
Pm =∑
a≤i1<i2<···<im≤n
P (Ai1 ∩ Ai2 · · · ∩ Aim) =(
nm
)
(n−m)!n!
= n!(h−m)!m!
(n−m)!n!
= 1m!
P (n) = 1 − 12!
+ 13!− 1
4!· · ·+ (−1)n+1 1
n!
Special Cases: Number of permutations of n things in which there is at least
1 match = P (n) · n!.
n = 3 P (n)n! = 6 ×(
1 − 12
+ 16
)
= 4
n = 4 P (n)n! = 24(
1 − 12
+ 16− 1
24
)
= 15
Problem: Given any n events, A1, A2, · · ·An prove that the probability ofexactly m ≤ n events occurring is
P = Pm −(
m + 1
m
)
Pm+1 +
(
m + 2
m
)
Pm+2 · · · ±(
n
m
)
Pn
where
Pk =∑
1≤i1<i2
, · · · ik ≤ nP (Ai1 ∩ Ai2 ∩ · · · ∩ Aik)
9
Good Example of Bayesian Inference
Sometimes the application of Bayes’ theorem may yield results that appearcounter-intuitive.
Example: A laboratory test is developed to detect mononucleosis (mono,for short). The probability that a person selected at random has mono is0.005. If a person has mono, 95% of the time he test will be positive. If aperson does not have mono, the test will be positive only 4% of the time.These circumstances are described by the binary channel shown in Figure 5.
e
e
-
-
@
@@
@@
@R0.96
0.95
person w/o mono
person w/ mono positive mono test
negative mono test
0.05
0.04
Fig. 5
What is the probability that a person has mono conditioned on the fact thathis test came out positive?
M = person has monoT = positive mono test
prior probabilities conditional probabilities
P (M)P (M)
==
0.0050.995
P (T |M) = 0.95P (T |M) = 0.04
10
Then, by Bayes’ theorems,
a posteriori probability
P (M |T ) =P (T |M)P (M)
P (T |M)P (M) + P (T |M)P (M)
=0.95 × 0.005
0.95 × 0.005 + 0.04 × 0.995
=0.00475
0.00475 + 0.0398
=0.00475
0.04455
= 0.107 !
Thus the test might give rise to too many false alarms. How to improve?Bring down the probability P (T |M) from 0.04. Improve the test.
A useful form of Bayes’ theorem is obtained by conditioning in more thanone event.
Let H := hypothesis (e.g. a disease event),Let E := evidence of data (e.g. image data event), andLet C := context (e.g. age group). Then,
P (H|E ∩ C) =P (E|H ∩ C) · P (H|C)
P (E|C)
To see this, observe that the r.h.s. above
=P (E ∩ H ∩ C)
P (H ∩ C)· P (H ∩ C)
P (C)· P (C)
P (E ∩ C)
=P (E ∩ H ∩ C)
P (E ∩ C)
=P (H ∩ (E ∩ C))
P (E ∩ C)
11
= P (H|E ∩ C)
= l.h.s
12
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ENEE 324H Lecture 6Inequalities
Estimating probabilities is necessary where analytic formulas are hard tofind. Finding good estimates (upper and lower bounds) is an art. But thereare some basic estimates derivable from first principles.
1. Markov inequality
Let X be a non-negative random variable. Let u denote the unit stepfunction
u(x) = 1 x > 00 x < 0
Let a > 0. Then it is easy to see that
u(X − a) ≤ X
a
Then
E (u(X − a)) ≤ E(X)
a
But
E (u(X − a)) = 0 · Pω : X(ω) < a + 1 · Pω : X(ω) > a= P (X > a)
Thus
P (X > a) ≤ E(X)
a
Remark (a) Since ω : X(ω) > a ⊆ ω : X(ω) > ait follows that
Pω : X(ω) > a ≤ Pω : X(ω) > a
≤ E(X)
a
1
Remark (b) If the assumption of non-negativity of X is not applicable,one can still write
Pω : |X(ω) − µ| > a ≤ E
( |X − µ|a
)
where µ ∈ lR is arbitrary and a > 0. This observation leads to the nextinequality.
( )E Y ( )E Y ( )E Yx xx
( )Y
p y
Figure 1: Chebyshev’s inequality estimates the tail-probability
( ( ) )P Y E Y
= area under density curve
marked by hatch lines
= tail probability
2. Chebyshev inequality
Let Y be any real-valued random variable.Let X = µ + (Y − E(Y ))2
Set a = δ2 for δ > 0.Then by Markov’s inequality,
P ((Y − E(Y ))2 > δ2) ≤ E ((Y − E(Y ))2)
δ2
equivalently,
P (|Y − E(Y )| > δ) ≤ V ar(Y )δ2
2
3. Convex functions and Jensen’s inequality
f : lR → lR is convex if
f(αx + (1 − α)y) ≤ αf(x) + (1 − α)f(y)for α ∈ [0, 1].
From this, it follows that the derivative f ′(x) (if it exists) is increasingwith x, and for any fixed x0, there exists a constant λ such that
f(x) > f(x0) + λ(x − x0)
The line with slope λ, passing through (x0, f(x0)) is called the support-ing line at the point (x0, f(0)) as in Figure 2.
f
0f x
0x x
Figure 2
Let x0 = E(X) for a random variable X. Then,
E(f(X)) > E(f(x0) + λ(X − x0))
= E(f(E(X))) + E(λ(X − E(X))
= f(E(X)) + λE(X) − λE(X)
= f(E(X))
3
Thus, for a convex function f ,
E(f(X)) > f(E(X))
4. Chernoff’s inequality
Let X be any random variable. Given ǫ > 0, define a new randomvariable dependent on X,
Yǫ = 1 if x > ǫ
0 if X < ǫ
For any t, it follows that
etX > etǫ Yǫ
Hence
E(etX) > E(etǫ Yǫ)
= etǫ E(Yǫ)
= etǫ P (X > ǫ)
(Here we assume existence of relevant expectations.) Hence,
P (X > ǫ) ≤ e−tǫ E(etX)
The free parameter t in the above inequality can be used to obtain atighter estimate,
P (X > ǫ) ≤ inft>0
e−tǫ E(etX)
Here“infimum” stands for the greatest lower bound.
4
Example: Suppose X is Gaussian with mean 0 and variance 1. Then thedensity of X is
PX(x) =1√2π
exp
(−x2
2
)
−∞ < x < ∞
E(etX) =
∫
∞
−∞
etx 1√2π
e−x2/2 dx
= et2
2
∫
∞
−∞
1√1π
e
(
tx−x2
2−
t2
2
)
dx
= et2/2
∫
∞
−∞
1√2π
e−(x−t)
2
2 dx
= et2/2
Then
P (x >∈) ≦ inft>0
e−tǫ+t2/2
= e−ǫ2/2
5
